Magnetic Field: Solenoid - Physics 2

An ideal solenoid of radius r has interior B. If radius doubles but N, L, I are fixed, what happens to B?
Answer: The same The field inside an ideal solenoid is B = mu0 N I / L, which depends only on the number of turns, current, and length, not on the radius. So doubling r changes nothing about B. The tempting mistake is to think the field spreads out over a larger area and weakens, but B is set by the winding density and current, not the cross-sectional area.
A student uses the known end-field result to argue: 'If I place two identical finite solenoids end-to-end, aligned coaxially with currents in the same direction, the field at the junction between them equals $B_0/2 + B_0/2 = B_0$ .' Under what condition is this argument most physically meaningful, and what subtlety limits its practical usefulness?
Answer: The argument is most useful when $L \gg R$ (long solenoid limit). In this limit, $B_0 = \mu_0 n I$ accurately describes the nearly uniform interior field, so the junction field approaching $B_0$ is physically meaningful. For short solenoids ( $L \sim R$ ), the interior field is non-uniform and $B_0$ no longer represents a well-defined interior value, so the superposition result, while mathematically correct, loses its practical significance. Whenever you see a superposition argument applied to solenoids, ask yourself: what physical regime makes the underlying formula meaningful? The end-field result $B_{end} = B_0/2$ is mathematically exact for any finite solenoid, but $B_0 = \mu_0 n I$ only represents a well-defined, uniform interior field when $L \gg R$ . C is correct because the student's superposition argument is mathematically valid in all cases — placing two solenoids end-to-end genuinely produces $B_0/2 + B_0/2 = B_0$ at the junction. The real question is whether $B_0$ is a meaningful benchmark. In the long solenoid limit ( $L \gg R$ ), the interior field is nearly uniform and well-approximated by $\mu_0 n I$ , so saying the junction recovers $B_0$ is physically informative: the seam essentially "heals" the edge effect. For short solenoids ( $L \sim R$ ), the field is non-uniform throughout and $B_0$ doesn't represent any characteristic interior value — so the result, while correct, tells you little. A is wrong because it ignores that "valid" and "physically meaningful" are different things. Superposition always works, but the significance of the result depends on context. B is wrong and reverses the physics entirely. The $B_0/2$ end-field result holds for any aspect ratio; short solenoids ( $L \ll R$ ) are precisely where $B_0$ becomes meaningless, not where the formula is uniquely valid. D is wrong because it invents a fictitious "back-reaction" mechanism. Radial fringe fields do exist but don't reduce the axial superposition — that's not how magnetostatics works. Study tip: On questions mixing superposition with solenoid physics, always separate mathematical correctness from physical significance — examiners love to test whether you know the difference.
An ideal solenoid has $n$ turns per unit length, radius $R$ , and carries current $I$ . A student argues that if the solenoid is stretched uniformly so that its length doubles while the wire is re-wound to maintain the same total number of turns $N$ , the magnetic field inside doubles because the solenoid is now longer. A second student argues the field is halved. A third student claims the field is unchanged. Which student is correct, and what is the correct field after stretching?
Answer: The second student is correct. Stretching doubles the length while $N$ is fixed, so the turns density drops to $n' = N/(2L) = n/2$ . Applying $B = \mu_0 n' I$ gives $B' = \mu_0(n/2)I = B/2$ , so the field is halved. Whenever you see a solenoid problem involving physical changes, your first instinct should be to return to the fundamental formula and track exactly which quantities change: $B = \mu_0 n I$ , where $n$ is the turns per unit length, not the total number of turns. Here's the key insight: when the solenoid is stretched so its length doubles from $L$ to $2L$ , the total number of turns $N$ stays fixed (the wire is re-wound to preserve $N$ ), but those turns are now spread over twice the length. The new turns density becomes $n' = N/(2L) = n/2$ . Plugging into Ampere's law, which for an ideal solenoid gives $B = \mu_0 n I$ , yields $B' = \mu_0 (n/2) I = B/2$ . The field is halved — the second student is correct, making B the right answer. Choice A contains a fundamental misconception: Ampere's law doesn't reward longer path length with a stronger field. A longer Amperian loop encloses more length but proportionally fewer turns per unit length, so the enclosed current per unit length actually decreases. Choice C is wrong because $B$ depends on $n$ (turns per unit length), not on $N$ alone — $\mu_0 N I$ isn't even dimensionally the correct formula for field. Choice D is a fabricated compensation argument; flux per turn and turns density are not related in a way that holds $B$ constant, and $n$ itself changes when you stretch the solenoid. Your study tip: always distinguish between $N$ (total turns) and $n$ (turns per unit length). Solenoid problems frequently exploit this difference — if the geometry changes, recompute $n$ before applying $B = \mu_0 n I$ .
Solenoid 1 has n turns/m and current I; solenoid 2 has 2n turns/m and current I/3. What is $B_2/B_1$ ?
Answer: 2/3 Solenoid field equals mu_0 times turns per meter times current. Solenoid 2 gives mu_0 times 2n times I/3, which is (2/3)mu_0 nI, so B2/B1 = 2/3. A tempting mistake is 3/2, but that flips the ratio and ignores that the one-third current outweighs the doubled turns.
Which of the following statements best evaluates the student's claim and identifies the correct expression for the magnetic field inside the toroid at radius $r$ (where $a < r < b$ )?
Answer: The claim is correct as a computational shortcut but physically incomplete: the field is $B = \mu_0 N I / (2\pi r)$ , and it **varies with $r$ ** inside the winding region, being strongest near $r = a$ and weakest near $r = b$ , unlike a straight solenoid where $B$ is uniform. Whenever you encounter a question about magnetic fields in a toroid, your anchor concept should be Ampère's Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$ . The key insight is that you choose a circular Amperian loop of radius $r$ centered on the toroid's axis, and symmetry guarantees $B$ is constant along that loop. For a circular path at radius $r$ (where $a < r < b$ ), the loop encloses all $N$ turns carrying current $I$ , so $I_{enc} = NI$ . Applying Ampère's Law: $B(2\pi r) = \mu_0 NI$ , which gives $B = \mu_0 NI / (2\pi r)$ . The student's shortcut of writing $n = N/(2\pi r)$ as a local turns-per-unit-length is mathematically equivalent and gives the right formula — but it obscures a critical physical truth: because $r$ appears in the denominator, the field is not uniform inside the winding. It is strongest near the inner radius $a$ and weakest near the outer radius $b$ . This makes B the correct answer. Choice A is wrong because it claims the field is uniform inside the toroid — it is not. Uniformity holds for an ideal straight solenoid, not a toroid, where the varying path circumference creates a radial dependence. Choice C is wrong because $\mu_0 NI \ln(b/a)/(2\pi)$ is actually the formula for the flux through the toroid's cross-section, not the field at a point. Choice D is wrong because using the mean circumference $\pi(a+b)$ gives only an average approximation, not the exact field at a specific $r$ . Your study tip: always distinguish between the field at a point (found via Ampère's Law with a loop through that point) and derived integrated quantities like flux. Confusing these is one of the most common traps in magnetostatics problems.
An ideal solenoid has 500 turns/m and current 2.0 A. Viewed from the left, current is clockwise. What is B inside?
Answer: 1.3 mT, rightward Inside a solenoid, B = mu0 n I = (4 pi x $10^{-7}$ )(500)(2) = 1.3 mT. For clockwise current viewed from the left, your right hand curls clockwise and your thumb points away from you, so the field inside points rightward. The tempting wrong choice is 1.3 mT leftward: the magnitude is right, but that direction would mean counterclockwise when viewed from the left.
An ideal solenoid with $n = 5000 \text{ turns/m}$ and radius $R = 2.0 \text{ cm}$ carries a current that varies in time as $I(t) = I_0 \sin(\omega t)$ where $I_0 = 4.0 \text{ A}$ and $\omega = 100\pi \text{ rad/s}$ . At time $t = 1/(200) \text{ s}$ , what is the magnitude of the magnetic field inside the solenoid?
Answer: $B = \mu_0 n I_0 = (4\pi\times10^{-7})(5000)(4.0) \approx 25.1 \text{ mT}$ , because $t = 1/200 \text{ s}$ corresponds to $\omega t = \pi/2$ , where $\sin(\pi/2) = 1$ , so the current is at its maximum value $I_0$ . When a solenoid carries a time-varying current, the magnetic field inside it is still given by the same static formula — $B = \mu_0 n I$ — evaluated at the instantaneous current. This is the quasi-static approximation, valid when the current changes slowly enough that we don't need to account for radiation effects (which is the case here). So the approach is straightforward: find $I$ at $t = 1/200$ s, then plug into $B = \mu_0 n I$ . Evaluating the argument: $\omega t = 100\pi \cdot \frac{1}{200} = \frac{\pi}{2}$ . Therefore $I = I_0 \sin(\pi/2) = I_0 = 4.0$ A. The field is $B = \mu_0 n I_0 = (4\pi \times 10^{-7})(5000)(4.0) \approx 25.1 \text{ mT}$ , confirming A is correct. Choice B confuses the magnetic field with the induced EMF. The EMF around a loop does depend on $dB/dt \propto dI/dt \propto \cos(\omega t)$ , but the field itself tracks $I(t)$ , not its derivative. Don't mix up Faraday's law with the field formula. Choice C invents a fictitious "EMF-driven field" proportional to $\omega$ . No such term exists in the quasi-static solenoid field. The field depends only on the instantaneous current, not on how fast it's changing. Choice D misapplies RMS. RMS current is useful for computing average power, not for finding the instantaneous magnetic field. The field at any moment equals $\mu_0 n I(t)$ , period. Study tip: Whenever you see a time-varying current in a solenoid, remember that $B = \mu_0 n I(t)$ — just substitute the instantaneous current. Reserve RMS for power calculations and $dI/dt$ for EMF calculations.
What is the magnitude of the net magnetic field in the interior region where both solenoids overlap, far from any end effects?
Answer: $B = \mu_0 |n_1 I_1 - n_2 I_2| = \mu_0 |3000 - 2000| = 1000\mu_0 \approx 1.26 \text{ mT}$ , because the fields are antiparallel and the net field is the difference of $n_1 I_1$ and $n_2 I_2$ . When two solenoids share an axis, you treat each one as an independent source of magnetic field and apply superposition — the net field is the vector sum of the individual fields. The key detail here is the winding direction, which determines whether each solenoid's field points along the same axis direction or opposite directions. For a single ideal solenoid, the interior field is $B = \mu_0 n I$ , directed along the axis according to the right-hand rule. Here, Solenoid 1 produces $B_1 = \mu_0 (1000)(3.0) = 3000\mu_0$ and Solenoid 2 produces $B_2 = \mu_0 (2000)(1.0) = 2000\mu_0$ . Because the solenoids are wound in opposite senses, their fields point in opposite directions inside the overlap region. Superposition then gives a net magnitude of $|B_1 - B_2| = \mu_0|3000 - 2000| = 1000\mu_0 \approx 1.26 \text{ mT}$ , confirming that A is correct. B is wrong because it ignores winding direction entirely — adding the magnitudes only applies when fields are parallel (same winding sense). C misapplies Lenz's law: that principle governs induced currents from changing flux, not the static superposition of steady fields. A steady solenoid field cannot "shield" anything by Lenz's law. D invents an averaging rule that has no physical basis; superposition adds fields at each point in space, not averages them by volume. A useful habit: whenever a problem mentions winding direction or current circulation, immediately assign a directional sign ( $+$ or $-$ ) to each field before doing any arithmetic. Opposite windings mean subtraction; same windings mean addition.
What is the magnitude of the magnetic field deep inside this double-wound solenoid, and which single-layer solenoid configuration is equivalent to it?
Answer: $B = \mu_0(n_1 I_1 + n_2 I_2) = \mu_0(3000 + 3000) = 6000\mu_0 \approx 7.54 \text{ mT}$ ; equivalent to a single solenoid with any $n$ and $I$ satisfying $nI = 6000 \text{ A/m}$ , such as $n = 3000 \text{ turns/m}$ and $I = 2.0 \text{ A}$ . When dealing with a multi-layer solenoid, the key principle is superposition: each layer independently creates its own magnetic field deep inside, and you simply add them. For an ideal solenoid, the field is $B = \mu_0 n I$ , where $n$ is turns per meter and $I$ is the current. Because both layers' currents circulate in the same direction, their fields point the same way and add constructively. Applying superposition here gives $B = \mu_0 n_1 I_1 + \mu_0 n_2 I_2 = \mu_0(1500)(2.0) + \mu_0(1000)(3.0) = \mu_0(3000 + 3000) = 6000\mu_0 \approx 7.54 \text{ mT}$ . Any single-layer solenoid that produces this same field only needs to satisfy $nI = 6000 \text{ A/m}$ — for instance, $n = 3000 \text{ turns/m}$ with $I = 2.0 \text{ A}$ , or countless other combinations. That's exactly what C states, making it correct. A gets the field calculation right but claims the only equivalent solenoid must use $n_1 + n_2$ with $I_1 + I_2$ . This is false — any $(n, I)$ pair satisfying $nI = 6000$ works; there's no unique equivalent configuration. B incorrectly uses the mean current rather than each layer's actual current. Averaging the currents has no physical basis here — you need to weight each layer's contribution by its own $n$ and $I$ . D introduces a fictitious "shielding" effect. Solenoid layers don't screen each other magnetically; both layers contribute fully to the interior field. Study tip: Whenever you see multiple current-carrying layers in a solenoid, immediately think superposition — calculate $\mu_0 n_i I_i$ for each layer separately and sum them. There's no shielding, no averaging, just addition.
An ideal solenoid with $n$ turns per unit length, radius $R$ , and current $I$ has a magnetic field $B = \mu_0 n I$ inside and zero outside. A student claims that if the solenoid is bent into a complete torus (a toroid) without changing $n$ , $R$ , or $I$ , the field inside the toroid will still equal $\mu_0 n I$ everywhere, since the local winding geometry is unchanged. Which response best evaluates this claim?
Answer: The claim is incorrect. Bending the solenoid into a toroid changes the effective turns density seen by an Amperian loop: inner turns are closer together (higher local $n$ ) and outer turns are farther apart (lower local $n$ ). The field varies as $B(r) = \mu_0 N I/(2\pi r)$ , where $n$ in the original solenoid is the turns density at the mean radius, and the field is not uniform. Whenever you see a question comparing a solenoid to a toroid, the key tool is Ampere's law applied globally, not just a local geometric argument. The law states $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$ , and the shape of your Amperian loop determines everything. For a toroid with $N$ total turns, choose a circular Amperian loop of radius $r$ centered on the toroid's axis. The enclosed current is $N I$ , and by symmetry $B$ is constant along the loop, giving $B(2\pi r) = \mu_0 N I$ , so $B(r) = \mu_0 N I / (2\pi r)$ . This field clearly depends on $r$ : it is stronger near the inner radius and weaker near the outer radius. The original solenoid's uniform $n$ becomes a non-uniform effective turns-per-length around the torus — inner windings are compressed (higher local density) and outer windings are stretched (lower local density) — making the field nonuniform. Answer B captures this exactly. Answer A is the trap the question is built around. Local winding geometry looks unchanged, but Ampere's law is a global statement. The curvature of the path means the same $N$ turns contribute differently at different radii, and you cannot ignore that by reasoning "locally." Answer C invents a fictional energy-based reduction mechanism. Toroidal geometry confines the field (which is actually an advantage), but confinement does not reduce field magnitude. Answer D is a plausible-sounding compromise, but it's wrong: $\mu_0 n I$ does not hold at the mean radius specifically, nor does the cross-section average equal $\mu_0 n I$ in general. Study tip: When geometry changes, always re-apply Ampere's law from scratch — local intuition can mislead you when path curvature matters.
An ideal solenoid has $N$ turns, length $L$ , cross-sectional radius $R$ , and carries current $I$ . The wire is now replaced with a superconducting wire of the same gauge, and the solenoid geometry is unchanged. The current is then slowly increased to $2I$ while the solenoid is maintained below its critical temperature. Simultaneously, the solenoid length is doubled to $2L$ by adding an equal number of turns (so total turns become $2N$ ) while keeping $R$ constant. What is the ratio of the new magnetic field to the original magnetic field?
Answer: $B_{new}/B_{old} = 2$ , because although the length doubles, the number of turns doubles as well, keeping $n$ constant, while the current doubles; these two effects multiply to give a net factor of 2. Whenever you see a solenoid problem with multiple simultaneous changes, your anchor should be the field formula $B = \mu_0 n I$ , where $n = N/L$ is the turn density (turns per unit length). Apply each change to this formula systematically before combining them. Here, the length doubles to $2L$ and the number of turns doubles to $2N$ , so the new turn density is $n_{new} = 2N/2L = N/L$ . The turn density is unchanged. However, the current doubles from $I$ to $2I$ . Plugging into the formula: $B_{new} = \mu_0 n_{new}(2I) = \mu_0 (N/L)(2I) = 2\mu_0 nI = 2B_{old}$ . The ratio is exactly 2, confirming answer A. Answer B claims the turn density doubles, but this is wrong — adding turns proportionally to the added length keeps $n$ constant. Doubling $N$ alone would double $n$ , but doubling both $N$ and $L$ together cancels out, leaving $n$ the same. The factor-of-4 result follows from a false premise. Answer C invokes flux conservation in a superconductor, but this only applies to a closed superconducting loop carrying a fixed, externally-unchanged current. Here, the current is being driven externally from $I$ to $2I$ , so flux conservation does not freeze the field. Answer D uses a fabricated formula $B \propto \sqrt{N \cdot I}$ , which has no physical basis. The correct dependence is linear in both $n$ and $I$ . Study tip: When a solenoid problem changes both $N$ and $L$ , always compute $n = N/L$ first — it often simplifies dramatically before you even touch the current.
An ideal solenoid is 5.0 cm long, has 200 turns, and carries 2.0 A. What is B inside?
Answer: 10 mT The field inside a solenoid is mu_0 n I. Convert length to meters: n = 200 / 0.050 = 4,000 turns/m. Then B = (4 pi x $10^{-7}$ )(4,000)(2.0) ≈ 0.010 T = 10 mT. A tempting wrong answer is 0.10 mT, which comes from leaving length in cm as turns/cm instead of converting to turns/m.
An ideal solenoid of length $L$ , radius $R$ , total turns $N$ , and current $I$ is oriented along the $z$ -axis. A long straight wire carrying current $I_0$ is placed along the central axis of the solenoid. Which of the following correctly describes the magnetic field inside the solenoid far from its ends?
Answer: The field inside is $\sqrt{(\mu_0 n I)^2 + (\mu_0 I_0 / 2\pi r)^2}$ , directed at an angle to the axis that depends on $r$ , and the net axial component equals $\mu_0 n I$ while the net azimuthal component equals $\mu_0 I_0/(2\pi r)$ at every interior point. When multiple current sources exist in the same region, you must apply superposition carefully — not just add magnitudes, but recognize that fields pointing in different directions combine as vectors, not scalars. Here, the solenoid produces a uniform axial field $B_{sol} = \mu_0 n I$ in the $\hat{z}$ direction throughout its interior (where $n = N/L$ ). The central wire produces a field $B_{wire} = \mu_0 I_0 / (2\pi r)$ that curls azimuthally in the $\hat{\phi}$ direction at every interior point a distance $r$ from the axis. These two fields are perpendicular to each other at every interior point — one axial, one azimuthal — so you cannot simply add their magnitudes. Instead, the net field magnitude is the vector sum: $B_{net} = \sqrt{(\mu_0 n I)^2 + (\mu_0 I_0 / 2\pi r)^2}$ , and its direction tilts at an angle that depends on $r$ . This makes D correct. A is wrong because it claims the field is "directed azimuthally near the axis and axially far from it," which mischaracterizes the geometry — both components exist simultaneously at every interior point, not in separate regions. B is the most tempting trap: it correctly identifies that the wire's field is azimuthal and the solenoid's is axial, but then wrongly concludes the wire's field can simply be ignored. Orthogonality means you add them as perpendicular vectors — the wire's field is real and measurable inside. C incorrectly evaluates the wire's field at $r = R$ (the wall) and treats it as a uniform axial addition, which is wrong on both counts — the wire field is azimuthal and varies with $r$ . Study tip: Whenever fields from two sources are perpendicular, always use $B = \sqrt{B_1^2 + B_2^2}$ — never $B_1 + B_2$ . Orthogonal doesn't mean ignorable.
An ideal solenoid is stretched to twice its length with same total turns and current. What happens to B?
Answer: Halved The magnetic field inside a solenoid is B = mu0 times (turns per length) times current. Stretching it to twice the length with the same total turns cuts the turns per length in half, so B is halved. The tempting wrong answer is unchanged, but the total number of turns doesn't set B; the turns per unit length does.

Start this quiz